Problem: Jeremiah makes $25\%$ of the three-point shots he attempts. For a warm up, Jeremiah likes to shoot three-point shots until he makes one. Let $M$ be the number of shots it takes Jeremiah to make his first three-point shot. Assume that the results of each shot are independent. Find the probability that it takes Jeremiah more than $6$ attempts to make his first shot. You may round your answer to the nearest hundredth. $P(M>6)=$
Answer: Without a fancy calculator On each shot: $P({\text{make}})=0.25$ $P(\text{miss}})=0.75$ If the first time he makes the shot occurs after his $6^{\text{th}}$ attempt, then he needs to miss his first $6$ shots. $\begin{aligned} P(M>6)&=P(\text{miss first 6}) \\\\ &=(0.75})^6 \\\\ &\approx0.178 \end{aligned}$ $P(M>6)\approx0.178 \approx 0.18$